Place a ruler against the curve at the required point and draw a straight line that touches the curve there without cutting it. Read two points off the tangent, not off the curve, then apply $\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}$.

\begin{align*} \text{Gradient} & = \frac{y_2 - y_1}{x_2 - x_1} \\ & = \frac{6 - (-4)}{2 - 0} \\ & = 5 \end{align*}

\begin{align*} y & = mx + c \\ y & = (-1)x + 0 \qquad [\text{Form a line with gradient -1 and } y \text{-intercept } 0] \\ y & = -x \end{align*}

Draw a tangent to the curve that is parallel to $y = -x$:

$$ \text{Point: } (-2, - 5) $$

\begin{align*} & \text{The line } y = - 6 \text{ does not meet the curve } y = x^2 + 3x - 3. \\ & \text{Thus, the equation } x^2 + 3x - 3 = -6 \text{ has no real solutions.} \end{align*}

\begin{align*} x^2 + 3x + 1 & = 0 \\ x^2 + 3x + 1 - 4 & = 0 - 4 \qquad [\text{Subtract } 4 \text{ on both sides}] \\ \underbrace{x^2 + 3x - 3}_\text{Curve} & = -4 \\ \\ \text{Draw } & y = -4 \end{align*}

$$ \text{From graph, } x = -2.6 \text{ or } - 0.4 $$

\begin{align*} 2x + 1 & = \frac{8}{x} \\ x(2x + 1) & = x \left( \frac{8}{x} \right) \qquad [\text{Multiply each term by } x ] \\ 2x^2 + x & = 8 \\ 2x^2 + x - 8 & = 0 \\ x^2 + 0.5x - 4 & = 0 \qquad \qquad \quad [\text{Divide each term by } 2 ] \\ x^2 + 0.5x + 2.5x - 4 & = 0 + 2.5x \\ x^2 + 3x - 4 + 1 & = 2.5x + 1 \\ \underbrace{x^2 + 3x - 3}_\text{Curve} & = 2.5x + 1 \\ \\ \text{Draw } & y = 2.5x + 1 \end{align*}

$$ \text{From graph, } x = -2.25 \text{ or } 1.75 $$

\begin{align*} \frac{1}{2}x^2 & \ge -2x \\ x^2 & \ge -4x \qquad [\text{Multiply every term by } 2] \\ x^2 + 4x & \ge 0 \\ x^2 + 4x - x & \ge 0 - x \\ x^2 + 3x & \ge - x \\ \underbrace{x^2 + 3x - 3}_\text{Curve} & \ge - x - 3 \\ \\ \text{Draw } & y = -x - 3 \end{align*}

(We require the parts of the curve that lie on or above the straight line - highlighted in yellow.)

$$ \text{From graph, } x \le -4 \text{ or } x \ge 0 $$

\begin{align*} y + x & = -1 \\ y & = -x - 1 \end{align*}

At the points where the line $ y = -x - 1$ intersects the curve $y = x^2 + 3x - 3$, their $y$-values are equal.

\begin{align*} - x - 1 & = x^2 + 3x - 3 \\ 0 & = x^2 + 3x + x - 3 + 1 \\ 0 & = x^2 + 4x - 2 \\ \\ \text{Comparing } & \text{with } x^2 + ax + b = 0, \\ \\ a & = 4, b = -2 \end{align*}